最大化产品

$$E[D1] = E[D2] = \frac{1+2+3+4+5+6}6 = 3.5$$ 由于两个卷是独立的，因此产品的期望值是期望值的乘积： $$E[D1\cdot{D2}] = 3.5 \cdot 3.5 = 12.25$$

Original Explanation

Since the two rolls are independent, the expected value of the product is the product of the expected values:

假设 $D1$ 是具有较高值的骰子。在最佳玩法下，我们永远不会重新掷两个骰子中的最大值。为了找到最大值的预期值，让我们计算一个值可能成为最大值的方式的数量。对于 $1$，你必须掷出两个 $1's$。对于 2 作为你的最大值，你只能滚动 $1$ 和 $2's$，这为你提供了 $2\cdot2 -1$ 方法。 （这包括 $(1,1)$ 情况，因此减去以避免重复计数）继续此过程以获得： $$E[D1] = \frac{ 1(1) + 3(2) + 5(3) + 7(4) + 9(5) + 11(6) }{36} = \frac{161}{36}$$ 然后我们可以使用类似的计算来获得 $D2$ 的期望值，即最小值。然而，如果 $D2$ 低于重新掷骰的预期值，我们会用 $3.5$ 替换该值。因此， $$E[D2] = \frac{ 27(3.5) + 5(4) + 3(5) + 1(6) }{36} = \frac{271}{72}$$ 因为这些骰子彼此独立，所以我们可以将期望相乘，得到： $$E[D1\cdot{D2}] = {\frac{161}{36}} \cdot \frac{271}{72}= \frac{43631}{2592} \simeq 16.83$$

import random
from typing import Callable

roll_dice: Callable[[None], int] = lambda: random.randrange(1, 7)

products = []

for i in range(1000000):

    d1 = roll_dice()
    d2 = roll_dice()

    product = d1 * d2

    if min(d1, d2) < 3.5:
        reroll = roll_dice()
        product = max(d1, d2) * reroll

    products.append(product)

#expecting something close to 16.83
print(sum(products) / len(products))

Original Explanation

Suppose $D1$ is the die with the higher value. With optimal play, we would never re-roll the maximum of the two dice. To find the expected value of the maximum, let's count the number of ways a value could be a maximum. For $1$, you must roll two $1's$. For 2 to be your maximum, you can only roll $1$'s and $2's$ which gives you $2\cdot2 -1$ ways. (This includes $(1,1)$ case so subtract to not double count) Continue this process to get:

We can then use a similar computation to get the expected value of $D2$, the minimum. However, in the case the $D2$ is below the expected value of a re-roll, we replace that value with the $3.5$. Thus,

Because these die rolls were indendent of each other, we can multiply the expectations resulting in:

import random
from typing import Callable

roll_dice: Callable[[None], int] = lambda: random.randrange(1, 7)

products = []

for i in range(1000000):

    d1 = roll_dice()
    d2 = roll_dice()

    product = d1 * d2

    if min(d1, d2) < 3.5:
        reroll = roll_dice()
        product = max(d1, d2) * reroll

    products.append(product)

#expecting something close to 16.83
print(sum(products) / len(products))